Question: $f(x)=\begin{cases} \dfrac{1}{x+1}&\text{for }-5<x\leq -4 \\\\ 2^x&\text{for }x>-4 \end{cases}$ Find $\lim_{x\to -4^+}f(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4$ (Choice B) B $-\dfrac13$ (Choice C) C $\dfrac{1}{16}$ (Choice D) D The limit doesn't exist.
Answer: Notice that we were asked to find the one-sided limit, $\lim_{x\to -4^+}f(x)$. This is the limit where $x$ -values approach $-4$ from the right. Let's find the limit as $x$ approaches $-4$ from the right. We will use the fact that $f(x)=2^x$ for $x$ -values greater than $-4$. $\begin{aligned} &\phantom{=}\lim_{x\to -4^+}f(x) \\\\ &=\lim_{x\to -4^+}2^x \\\\ &=2^{-4}&\gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{16} \end{aligned}$ In conclusion, we found that $\lim_{x\to -4^+}f(x)=\dfrac{1}{16}$.